Question

Question:Prove that for any set

$\left(A-B\right)\mathbf{\cup }\left(B-A\right)\mathbf{=}\left(A\cup B\right)\mathbf{-}\left(A\cap B\right)$

NOTE:$\mathrm{\mathbb{Z}}$is the set of integers,$\mathrm{\mathbb{Q}}$is the set of rational numbers and $\mathrm{ℝ}$the set of real numbers.

Short answer

Answer:

It is proved that $\left(A-B\right)\cup \left(B-A\right)=\left(A\cup B\right)-\left(A\cap B\right)$.

Step-by-step solution

Describe the De-Morgan’s law 

According to theDe-Morgan’s law,

$\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{\cap }{\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{\cap }\mathbf{(}\mathbf{A}\mathbf{\text{'}}\mathbf{\cup }\mathbf{B}\mathbf{\text{'}}\mathbf{)}$

Step 2: Prove that  A-B∪B-A=A∪B-A∩B

The statement to be proved is,

$\left(A-B\right)\cup \left(B-A\right)=\left(A\cup B\right)-\left(A\cap B\right)$

Apply the definition of minus in LHS,

$\left(A-B\right)\cup \left(B-A\right)=\left(A\cup B\right)\cap {\left(A\cap B\right)}^{\text{'}}$

Apply De-Morgan’s law

$\left(A\cup B\right)\cap {\left(A\cap B\right)}^{\text{'}}=\left(A\cup B\right)\cap \left(A\text{'}\cup B\text{'}\right)$

Apply the intersection distributes over the union,

$$\begin{gathered} \left[\left(A\cup B\right)\cap {\left(A\cap B\right)}^{\text{'}}\right]=\left[\left(A\cup B\right)\cap A\text{'}\right]\cup \left[\left(A\cup B\right)\cap B\text{'}\right] \\ =\left[\left[\left(A\cap A\text{'}\right)\cup \left(B\cap A\text{'}\right)\right]\cup \left[\left[\left(A\cap B\text{'}\right)\cup \left(B\cap B\text{'}\right)\right]\right]\right] \end{gathered}$$

Now the intersection of a set with its compliment is empty.

$$\begin{gathered} \left[\left[\left(A\cap A\text{'}\right)\cup \left(B\cap A\text{'}\right)\right]\cup \left[\left[\left(A\cap B\text{'}\right)\cup \left(B\cap B\text{'}\right)\right]\right]\right]=\left(\left[\varphi \cup \left(B\cap A\text{'}\right)\right]\cup \left[A\cap B\text{'}\cup \varphi \right]\right) \\ =\left(B\cap A\text{'}\right)\cup \left(A\cap B\text{'}\right) \end{gathered}$$

Apply the definition of set minus,

$$\begin{gathered} \left(B\cap A\text{'}\right)\cup \left(A\cap B\text{'}\right)=\left(B-A\right)\cup \left(A-B\right) \\ =\left(A-B\right)\cup \left(B-A\right) \end{gathered}$$

Hence, the statement is proved.